Problem: For integers $a$, $b$, $c$, and $d$, $(x^2+ax+b)(x^2+cx+d)=x^4+x^3-2x^2+17x-5$. What is the value of $a+b+c+d$?
Solution: We expand the expression on the left and attempt to match up the coefficients with those in the expression on the right. \begin{align*}
(x^2+ax+b)(x^2+cx+d) = x^4+cx^3 \ +& \ dx^2 \\
ax^3 \ +& \ acx^2+adx \\
\ +& \ \ bx^2 \ +bcx+bd
\end{align*} $$=x^4+x^3-2x^2+17x-5$$ So we have $a+c=1$, $ac+b+d=-2$, $ad+bc=17$, $bd=-5$.

From the final equation, we know that either $b=1, d=-5$ or $b=-1, d=5$. We test each case:

If $b=1, d=-5$, then $ac+b+d=ac-4=-2$, so $ac=2$. We substitute $a=1-c$ from the first equation to get the quadratic $c^2-c+2=0$. This equation does not have any integer solutions, as we can test by finding that the discriminant is less than zero, $(-1)^2-4(1)(2)=-7$.

If $b=-1, d=5$, then $ac+b+d=ac+4=-2$, so $ac=-6$. We substitute $a=1-c$ from the first equation to get the quadratic $c^2-c-6=0$, which has solutions $c=-2$ (so $a=3$) or $c=3$ (so $a=-2$). In either case, we get that $a+b+c+d=\boxed{5}$.

The remaining equation, $ad + bc = 17$, tells us that the coefficients are $a = 3, b = -1, c = -2, d = 5.$